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tir0nel
09-04-2008, 05:59 PM
MY teacher requires us to do all the derivatives the long way, but if i could just see one problem like this worked i could understand it.....

Find the f '(a)

F(T)= 2T+1 / T+3

The answer should be 5 / (a+3)^2

....if someone could explain this to me i would greatly appreciate it.

HallsofIvy
09-05-2008, 10:08 AM
MY teacher requires us to do all the derivatives the long way, but if i could just see one problem like this worked i could understand it.....

Find the f '(a)

F(T)= 2T+1 / T+3

The answer should be 5 / (a+3)^2

....if someone could explain this to me i would greatly appreciate it.
To start with it is just algebra: you want to form the "difference quotient" (f(T+h)- f(T))/h (the "average rate of change" of f between T and T+ h).

Since f(T)= (2T+1)/(T+ 3), f(T+h)= (2(T+h)+ 1)/((T+h)+ 3)= (2T+2h+1)/(T+3+h)

Now subtract: f(T+h)- f(T)= (2T+2h+1)/(T+3+h)- (2T+1)/(T+ 3). To subtract those fractions, get the common denominator which, here, is (T+3+h)(T+3). To get that denominator multiply numerator and denominator of the first fraction by T+ 3 and of the second fraction by T+ 3+ h.

The numerators are
(2T+2h+1)(T+ 3)= 2T^2+ 2hT+ T+ 6T+ 6h+ 3= 2T^2+ 7T+ 3+ 2hT+ 6h
and
(2T+1)(T+ 3+h)= 2T^2+ T+ 6T+ 3+ 2hT+ h= 2T^2+ 7T+2hT+ 3+ h

Subtracting 2T^2+ 7T+ 3+ 2hT+ 6h- (2T^2+ 7T+2hT+ 3+ h)= 5h.
Notice that almost everything canceled. Everything involving NOT involving h MUST cancel and certainly will as long as the function is continuous.
We might as well leave the denominator as (T+3+h)(T+ 3):

f(T+h)- f(T)= 5h/(T+3+h)(T+3)

To form the difference QUOTIENT we must divide by h which, here, just cancels the "h" in the numerator:
(f(T+h)- f(T))/h= 5/(T+3+h)(T+3).

All that is basic (though a bit complicated) algebra. The "calculus" part is the limit we take to get not the "average" rate of change between, say, T= a and T= a+h, but the rate of change AT T= a. Taking the limit as h goes to 0, since neither numerator nor denominator goes to 0 here is just
5/(T+ 3+ 0)(T+ 3)= 5/(T+3)^2. In particular, at T= a, that is 5/(a+3)^2.